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Sketch the region enclosed by the curves. Decide whether to integrate with respect to x or y. Then find the area of the region between $$x=9-{y^2}$$ and $$x={y^2}-9$$

What I have done so far is put the equations in terms of y and make them equal to each other. I got $$9-x=x+9$$ which goes to $$2x=0$$. I am unsure of how to find the bounds of the integral from this. Can anyone help?

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You should definitely integrate with respect to $y$. – 1015 Jan 28 '13 at 20:51
I did that however I am having trouble finding the bounds of the integral. – Gabrielle Jan 28 '13 at 20:56
@Gabrielle The community frowns upon questions copied verbatim from a textbook with no apparent effort on your part. If you include in your question the comment that you have decided how to integrate but cannot determine the bounds, the community will more likely to help. – Austin Mohr Jan 28 '13 at 20:59
Oh okay let me edit in what I have so far then. I'm just reviewing for a test and didn't know. :( – Gabrielle Jan 28 '13 at 21:00
@Gabrielle That's alright. I didn't want you to think that calculus questions are unwelcome here. We are generally happy to answer anything, but we want to be confident that you are taking the time to think about the problem first. Welcome to the community. – Austin Mohr Jan 28 '13 at 21:19
up vote 4 down vote accepted

You need to find the intersections of the two parabola. These occur when $9-y^2=y^2-9$, which is equivalent to $y^2-9=0$. So you have two intersections, namely $(0,-3)$ and $(0,3)$.

Draw a picture.

The area is given by $$ \int_{-3}^3(9-y^2-(y^2-9))dy=2\int_0^3(18-2y^2)dy. $$

I guess you can finish the calculation.

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Ohh! Thank you so much I was putting the equations to find the bounds in terms of x. That was the problem thank you ! – Gabrielle Jan 28 '13 at 21:05
No problem. Good luck with the test. – 1015 Jan 28 '13 at 21:06

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